Theorem dp15lemh 1161

Description: Part of proof (1)=>(5) in Day/Pickering 1982. (Contributed by NM, 2-Apr-2012.)

Hypotheses

Ref	Expression
dp15lema.1	d = (a2 ∪ (a0 ∩ (a1 ∪ b1)))
dp15lema.2	p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
dp15lema.3	e = (b0 ∩ (a0 ∪ p0))
dp15lemg.4	c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
dp15lemg.5	c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))

Assertion

Ref	Expression
dp15lemh	((a0 ∪ a1) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b1)) ≤ ((c0 ∪ c1) ∪ (b1 ∩ (a0 ∪ a1)))

Proof of Theorem dp15lemh

Step	Hyp	Ref	Expression
1		dp15lema.1	. . . . . 6 d = (a2 ∪ (a0 ∩ (a1 ∪ b1)))
2		dp15lema.2	. . . . . 6 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
3		dp15lema.3	. . . . . 6 e = (b0 ∩ (a0 ∪ p0))
4	1, 2, 3	dp15lemc 1156	. . . . 5 ((a0 ∪ a1) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b1)) ≤ (((a0 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ (b1 ∪ b2)))
5	1, 2, 3	dp15lemd 1157	. . . . 5 (((a0 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ (b1 ∪ b2))) = (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) ∩ (b1 ∪ b2)))
6	4, 5	lbtr 139	. . . 4 ((a0 ∪ a1) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b1)) ≤ (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) ∩ (b1 ∪ b2)))
7	1, 2, 3	dp15leme 1158	. . . 4 (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) ∩ (b1 ∪ b2))) ≤ (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1))) ∩ (b1 ∪ b2)))
8	6, 7	letr 137	. . 3 ((a0 ∪ a1) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b1)) ≤ (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1))) ∩ (b1 ∪ b2)))
9	1, 2, 3	dp15lemf 1159	. . 3 (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1))) ∩ (b1 ∪ b2))) ≤ (((a1 ∪ a2) ∩ (b1 ∪ b2)) ∪ (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1))))
10	8, 9	letr 137	. 2 ((a0 ∪ a1) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b1)) ≤ (((a1 ∪ a2) ∩ (b1 ∪ b2)) ∪ (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1))))
11		dp15lemg.4	. . 3 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
12		dp15lemg.5	. . 3 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
13	1, 2, 3, 11, 12	dp15lemg 1160	. 2 (((a1 ∪ a2) ∩ (b1 ∪ b2)) ∪ (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1)))) = ((c0 ∪ c1) ∪ (b1 ∩ (a0 ∪ a1)))
14	10, 13	lbtr 139	1 ((a0 ∪ a1) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b1)) ≤ ((c0 ∪ c1) ∪ (b1 ∩ (a0 ∪ a1)))

Syntax hints:   = wb 1   ≤ wle 2   ∪ wo 6   ∩ wa 7

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-ml 1122  ax-arg 1153

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131

This theorem is referenced by:  dp15  1162